JEE Main 2024MathematicsDefinite IntegrationMediumMCQ

JEE Main 2024 — Definite Integration Question with Solution

JEE Main 2024 (01 Feb Shift 1)

Question

The value of the integral ∫0π4xdxsin42x+cos42x equals:

Choose an option

Show full solutionCorrect option: C
Correct answer
C2π232

Step-by-step explanation

Let, I=∫0π4xdxsin42x+cos42x

Let, 2x=t then dx=12dt

⇒I=14∫0π2tdtsin4t+cos4t

⇒I=14∫0π2π2−tdtsin4π2−t+cos4π2−t

⇒I=14∫0π2π2dtsin4t+cos4t−I

⇒2I=π8∫0π2dtsin4t+cos4t

⇒2I=π8∫0π2sec4tdttan4t+1

Let, tant=y then sec2tdt=dy

⇒2I=π8∫0∞1+y2dy1+y4

⇒I=π16∫0∞1+1y2y2+1y2dy

Putting, y−1y=p

⇒I=π16∫−∞∞dpp2+22

⇒I=π162tan−1p2−∞∞

⇒I=π2162

⇒I=2π232

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About this question

This is a previous-year question from JEE Main 2024, covering the Definite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.